[leetcode] 206. Reverse Linked List

problem

code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        
        if (head == null  || head.next == null) return head;
        
        ListNode prev = null;
        while (head != null) 
        {            
                        
            /*
            head: head -> (val = 1, next=head2)  // head2 is origin head.next;
            
            1
            temp: HEAD.next -> head2(val = 2, next=head3)
            HEAD.next(head2): prev(null)
            
            * head1(HEAD) => (val = 1, next=null)
            
            prev: head -> head1(val = 1, next = null)
            head: temp: HEAD.next -> head2(val = 2, next = head3(original) )
            
            2
            temp: HEAD.next.next -> head3(val = 3, next=head4) // head3 is origin head.next.next;
            head2.next(head3): prev: head -> head1(val = 1, next = null)
            
            * head2(HEAD.next) => head1(val=1, next=null);
            
            prev: head.next -> head2(val=2, next=head1)
            head: head3(val=3, next=head4)
            
            3
            temp: HEAD.next.next.next -> head4(val = 4, next=head5)
            head3.next(head4) -> prev: head2(val=2, next=head1)
            
            * head3(HEAD.next.next) => head3(val=3, next=head2);
            
            prev: head.next.next -> head3(val=3, next=head2);
            head: head4(val=4, next=head5);
            
            4
            temp: HEAD.next.next.next.next -> head5(val = 5, next=null)
            head4.next(head5) -> prev: head3(val=3, next=head2)
            
            * head4(HEAD.next.next.next) => head4(val=4, next=head3);
            
            prev: head.next.next.next -> head4(val=4, next=head3);
            head: head5(val=5, next=null);
            
            5
            temp: HEAD.next.next.next.next.next -> head6(null)
            head5.next(head6) -> prev: head4(val=4, next=head3)
            
            * head5(HEAD.next.next.next.next) => head5(val=5, next=head4);
            
            prev: head.next.next.next.next -> head5(val=5, next=head4);
            head: head6(null); 
            
            head: null => false: escape;
            */
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }

        return prev;
        
    }
}

Time: O(N)
Space: O(1), constant